122.MET March 2021 Q.7(b)

Find the synchronous impedance and reactance of an alternator in which a given field current produces an armature current of 200 A on short-circuit and a generated e.m.f. of 50 V on open- circuit. The armature resistance is 0.1 ohm. To what induced voltage must the alternator be excited if it is to deliver a load of 100 A at a p.f. a of 0,8 lagging, with a terminal voltage of 200 V

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Given

Open circuit voltage, E1 = 50 V

 Short circuit current, I1 = 200 A

Armature reactance, Ra = 0.1 Ω

Current, I = 100 A

cos φ = 0.8

Terminal Voltage, V = 200 V

To find

Induced voltage, E0

Solution

The determination of voltage regulation in an alternator can be found out by using 3 methods.

1.Synchronous impedance method

2. The ampere turn method

3.Zero power factor method.

In this question we are using Synchronous impedance method. The vector diagram in this method is given by the diagram given below.

We know that Synchronous impedance, Zs = O.C voltage / S.C current

                                                                     = E1 / I1 

                                                                     = 50 / 200

                                                                     = 0.25 Ω

Synchronous reactance Xs = √ (Zs² – Ra²)

                                            = √ (0.25² – 0.1²)

                                            = 0.23 Ω

Now IRa = 100 x 0.1 = 10 V

Now IXs = 100 x 0.23 = 23 V

It is given that cos φ = 0.8 , So sin φ = 0.6

We know that,

E0 = √ (200 x 0.8 + 10)² + (200 x 0.6 + 23)²

       ^{= 222 V}

And the vector diagram can be drawn